奇异系统有着广泛的实际应用背景, 如在化工、航空航天、生物医学、电力网络和社会经济等领域.随着现代控制理论与控制方法越来越多地应用到工程系统，用奇异系统来描述与刻画实际控制系统，相较于正常系统能更好地描述实际系统.因而，奇异系统获得了广泛研究，取得不少成果^[1-3].

关于奇异系统线性二次最优控制问题，经过近几十年研究，在理论及应用方面都有大量的成果^[4-6].然而，由于在现实中往往存在着不可忽视的噪声^[7-10]，因此，随机奇异系统引起了不少学者的兴趣.Balasubramaniam等^[11]用神经网络法得到了随机奇异系统不定线性二次最优控制的Riccati微分方程.Nallasamy等^[12]用遗传编程法研究了随机线性奇异Takagi-Sugeno模糊时滞系统的线性二次最优控制问题.Zhang等^[13]研究了随机奇异系统的稳定性以及线性二次最优控制，得到了随机奇异系统均方容许的条件以及有限时域和无限时域线性二次最优控制的Riccati方程.

另一方面，由于微分博弈在很多情况下反映了决策者的理性思维方式，在经济、社会等领域有着广泛的应用，众多学者对这类问题表现出了持久的研究热情，对于确定性系统的微分博弈问题已经被很多人研究过^[14-15].在过去的几十年中，奇异系统的微分博弈由于其能很好地模拟现实系统中的实际情况，从而得到了广泛研究.其中，大多数学者研究的是确定型线性奇异系统的微分博弈问题^[16-18].

但据作者所知，目前关于随机奇异系统非零和微分博弈问题的研究还未见报道.基于此，本文讨论噪声依赖于状态和控制的随机奇异系统的非零和微分博弈问题，旨在丰富现有一般随机系统非零和博弈的理论.

为简化叙述，本文引入下述记号：Rⁿ为n维欧氏空间；A′表示矩阵A的转置；A^-1表示矩阵A的逆矩阵；S^n×n为全体n×n阶实对称矩阵构成的集合；Sⁿ为所有n阶对称矩阵; S₊ⁿ为Sⁿ的非负定矩阵；rank(A)为矩阵A的秩；deg(det(sI-A))为行列式sI- A的次数.

1 预备知识

考虑以下随机奇异系统

$ \left\{ \begin{array}{l} \mathit{\boldsymbol{E}}{\rm{d}}x\left( t \right) = \mathit{\boldsymbol{A}}x\left( t \right){\rm{d}}t + \mathit{\boldsymbol{C}}x\left( t \right){\rm{d}}w\left( t \right),\\ x\left( 0 \right) = {x_0},t \ge 0. \end{array} \right. $

(1)

其中x₀∈Rⁿ是系统的初始状态，A∈R^n×n，C∈R^n×n是状态变量x(t)的系数矩阵，E是给定的奇异矩阵，rank(E) < n, w(t) ∈R是在给定的完备概率空间(Ω, F, {F_t}_t≥0, P)上的一维标准Wiener过程，表示系统噪声.

定义1^[19]

1) 如果det(sE-A)不恒为0，则系统(1)是正则的.

2) 如果deg(det(sE-A)) =rank(E)，则系统(1)是无脉冲的.

3) 如果$\mathop {\lim }\limits_{t \to \infty } \varepsilon {\left| {x\left( t \right)} \right|^2} = 0 $，则系统(1)是均方稳定的.

4) 如果系统(1)是正则的、无脉冲的且均方稳定的，则系统(1)是均方容许的.

引理1^[13]  系统(1)是均方容许的充分条件是存在一个非奇异矩阵G，使得式(2)成立.

$ \left\{ \begin{array}{l} \mathit{\boldsymbol{E'G}} = \mathit{\boldsymbol{G'E}} \ge 0,\\ \mathit{\boldsymbol{A'G}} + \mathit{\boldsymbol{G'A}} + \mathit{\boldsymbol{C'E'GC}} < 0. \end{array} \right. $

(2)

证明  证明过程可参考文献[13].

2 问题描述

设(Ω, F, {F_t}_{t ≥0}, P)是一个给定的完备概率空间, 其上定义了一维标准Wiener过程{w(t)}_{t ≥0}，笔者考虑如下随机奇异系统

$ \left\{ \begin{array}{l} \mathit{\boldsymbol{E}}{\rm{d}}x\left( t \right) = \left[ {\mathit{\boldsymbol{A}}x\left( t \right) + {\mathit{\boldsymbol{B}}_1}{u_1}\left( t \right) + {\mathit{\boldsymbol{B}}_2}{u_2}\left( t \right)} \right]{\rm{d}}t + \\ \;\;\;\;\left[ {\mathit{\boldsymbol{C}}x\left( t \right) + {\mathit{\boldsymbol{D}}_1}{u_1}\left( t \right) + {\mathit{\boldsymbol{D}}_2}{u_2}\left( t \right)} \right]{\rm{d}}w\left( t \right),\\ x\left( 0 \right) = {x_0},t \in \left[ {0,T} \right]. \end{array} \right. $

(3)

其中，x(t)∈Rⁿ是状态变量，u₁(t)∈R^m和u₂(t)∈R^l是两博弈人的决策控制变量, 系数矩阵A，B₁，B₂，C，D₁，D₂为具有相应维数的常数矩阵.实矩阵E是奇异的，假定rank(E) < n.

对于每一个博弈人，定义二次型性能指标J₁(u₁, u₂; x₀, 0)和J₂(u₁, u₂; x₀, 0)：

$ \begin{array}{l} {J_1}\left( {{u_1},{u_2};{x_0},0} \right) = \varepsilon \left\{ {\frac{1}{2}\int_0^T {\left[ {x'\left( t \right){\mathit{\boldsymbol{Q}}_1}x\left( t \right) + } \right.} } \right.\\ \left. {{{u'}_1}\left( t \right){\mathit{\boldsymbol{R}}_{11}}{u_1}\left( t \right) + {{u'}_2}\left( t \right){\mathit{\boldsymbol{R}}_{12}}{u_2}\left( t \right)} \right]{\rm{d}}t + \\ \left. {\frac{1}{2}x'\left( T \right){\mathit{\boldsymbol{H}}_1}x\left( T \right)} \right\},\\ {J_2}\left( {{u_1},{u_2};{x_0},0} \right) = \varepsilon \left\{ {\frac{1}{2}\int_0^T {\left[ {x'\left( t \right){\mathit{\boldsymbol{Q}}_2}x\left( t \right) + } \right.} } \right.\\ \left. {{{u'}_1}\left( t \right){\mathit{\boldsymbol{R}}_{21}}{u_1}\left( t \right) + {{u'}_2}\left( t \right){\mathit{\boldsymbol{R}}_{22}}{u_2}\left( t \right)} \right]{\rm{d}}t + \\ \left. {\frac{1}{2}x'\left( T \right){\mathit{\boldsymbol{H}}_2}x\left( T \right)} \right\}. \end{array} $

(4)

其中，Q_τ、R_τ1、R_τ2和H_τ ≥0(τ =1, 2)都是适当维数的对称矩阵.

本文中，笔者将研究限定在两博弈人的控制策略均为线性状态反馈情形，即u₁(t)=K₁(t)x(t)∈Γ₁，u₂(t)=K₂(t)x(t)∈Γ₂，其中K_τ (τ =1, 2)是矩阵值函数.Γ₁和Γ₂表示u₁(t)和u₂(t)的策略空间.

定义2  线性反馈策略(u₁(t), u₂(t))∈Γ₁^a×Γ₂^a⊂Γ₁×Γ₂称为容许策略，假如其对应的闭环系统是无脉冲的.相应地，Γ₁^a×Γ₂^a⊂Γ₁×Γ₂称为容许策略空间.

问题  给定式(3)描述的随机奇异系统，寻找可行控制(u₁^*(t), u₂^*(t))∈Γ₁^a×Γ₂^a，使式(5)对于所有(u₁^*(t), u₂^*(t))∈Γ₁^a×Γ₂^a，(u₁ (t), u₂^* (t))∈Γ₁^a ×Γ₂^a，(u₁^*(t), u₂(t))∈Γ₁^a×Γ₂^a都成立.

$ \left\{ \begin{array}{l} {J_1}\left( {u_1^ * ,u_2^ * ;{x_0},0} \right) \le {J_1}\left( {{u_1},u_2^ * ;{x_0},0} \right),\\ {J_2}\left( {u_1^ * ,u_2^ * ;{x_0},0} \right) \le {J_2}\left( {u_1^ * ,{u_2};{x_0},0} \right). \end{array} \right. $

(5)

3 单人博弈

在这一部分，笔者将介绍两人博弈的退化情形——单人博弈，即随机线性二次最优控制问题，在此种情况下所得到的相关结论将为两人随机非零和博弈的求解奠定基础.

考虑如下受控连续随机奇异系统

$ \begin{array}{l} \mathit{\boldsymbol{E}}dx\left( t \right) = \left[ {\mathit{\boldsymbol{A}}x\left( t \right) + \mathit{\boldsymbol{B}}u\left( t \right)} \right]{\rm{d}}t + \left[ {\mathit{\boldsymbol{C}}x\left( t \right) + } \right.\\ \left. {\mathit{\boldsymbol{D}}u\left( t \right)} \right]{\rm{d}}w\left( t \right), \end{array} $

(6)

其中u(t)=K(t)x(t)∈ R^m.

性能指标函数为

$ \begin{array}{l} J\left( {u;{x_0},0} \right) = \varepsilon \left\{ {\frac{1}{2}\int_0^T {\left[ {x'\left( t \right)\mathit{\boldsymbol{Q}}x\left( t \right) + } \right.} } \right.\\ \left. {\left. {\left. {u'\left( t \right)\mathit{\boldsymbol{R}}u\left( t \right)} \right]{\rm{d}}t + \frac{1}{2}x'\left( T \right)\mathit{\boldsymbol{H}}x\left( T \right)} \right]} \right\}. \end{array} $

(7)

其中，Q、H、R都是对称矩阵，H≥0.

关于随机奇异系统的线性二次最优控制，文献[13]中有详细的叙述.下面，引用文献[13]中关于随机奇异系统最优控制的结果.该结果将为接下来的求解奠定基础.

引理1^[13]     考虑系统(6)，若以下Riccati微分方程

$ \left\{ \begin{array}{l} \mathit{\boldsymbol{E'}}\dot P\left( t \right)\mathit{\boldsymbol{E + E'}}P\left( t \right)\mathit{\boldsymbol{A + A'}}P\left( t \right)\mathit{\boldsymbol{E}} + \mathit{\boldsymbol{C'E'}}P\left( t \right)\mathit{\boldsymbol{EC}} + \\ \;\;\;K'\left( t \right)\left( {\mathit{\boldsymbol{B'}}P\left( t \right)\mathit{\boldsymbol{E}} + \mathit{\boldsymbol{D'E'}}P\left( t \right)\mathit{\boldsymbol{EC}}} \right) + Q = 0,\\ \mathit{\boldsymbol{E'}}P\left( T \right)\mathit{\boldsymbol{E}} = \mathit{\boldsymbol{H}},\\ K\left( t \right) = - {\left( {\mathit{\boldsymbol{R}} + \mathit{\boldsymbol{D'E'}}P\left( t \right)\mathit{\boldsymbol{ED}}} \right)^{ - 1}} \times \left( {\mathit{\boldsymbol{B'}}P\left( t \right)\mathit{\boldsymbol{E}} + } \right.\\ \;\;\;\;\left. {\mathit{\boldsymbol{D'E'}}P\left( t \right)\mathit{\boldsymbol{EC}}} \right),\\ \mathit{\boldsymbol{R}} + \mathit{\boldsymbol{D'E'}}P\left( t \right)\mathit{\boldsymbol{ED}} > 0. \end{array} \right. $

(8)

存在解P(t)∈ S₊ⁿ，则最优控制策略u^*(·)存在：

$ \begin{array}{l} {u^ * }\left( t \right) = K\left( t \right)x\left( t \right) = - {\left( {\mathit{\boldsymbol{R}} + \mathit{\boldsymbol{D'E'}}P\left( t \right)\mathit{\boldsymbol{ED}}} \right)^{ - 1}} \times \\ \left( {\mathit{\boldsymbol{B'}}P\left( t \right)\mathit{\boldsymbol{E}} + \mathit{\boldsymbol{D'E'}}P\left( t \right)\mathit{\boldsymbol{EC}}} \right)x\left( t \right), \end{array} $

(9)

且最优值为

$ J\left( {{u^ * };{x_0},0} \right) = \frac{1}{2}{{x'}_0}\mathit{\boldsymbol{E'}}P\left( 0 \right)\mathit{\boldsymbol{E}}{x_0}. $

(10)

4 主要结果

借助单人博弈情形的相关结论，可得到下述定理1.

定理1  考虑系统(3)~ (5)，若以下耦合Riccati微分方程组对于所有的(u₁(t), u₂(t))∈Γ₁^a×Γ₂^a, 存在解(P₁(t), P₂(t))∈ S₊ⁿ× S₊ⁿ

$ \left\{ \begin{array}{l} \mathit{\boldsymbol{E'}}{{\dot P}_1}\left( t \right)\mathit{\boldsymbol{E + E'}}{P_1}\left( t \right)\left( {\mathit{\boldsymbol{A}} + {\mathit{\boldsymbol{B}}_2}{K_2}\left( t \right)} \right) + \left( {\mathit{\boldsymbol{A}} + } \right.\\ \;\;\;\;{\left. {{\mathit{\boldsymbol{B}}_2}{K_2}\left( t \right)} \right)^\prime }{P_1}\left( t \right)\mathit{\boldsymbol{E}} + {\mathit{\boldsymbol{Q}}_1} + {{K'}_2}\left( t \right){\mathit{\boldsymbol{R}}_{12}}{K_2}\left( t \right) + \\ \;\;\;\;{\left( {\mathit{\boldsymbol{C}} + {\mathit{\boldsymbol{D}}_2}{K_2}\left( t \right)} \right)^\prime }\mathit{\boldsymbol{E'}}{P_1}\left( t \right)\mathit{\boldsymbol{E}}\left( {\mathit{\boldsymbol{C}} + {\mathit{\boldsymbol{D}}_2}{K_2}\left( t \right)} \right) + \\ {{K'}_1}\left( t \right)\left[ {{{\mathit{\boldsymbol{B'}}}_1}{P_1}\left( t \right)\mathit{\boldsymbol{E}} + {{\mathit{\boldsymbol{D'}}}_1}\mathit{\boldsymbol{E'}}{P_1}\left( t \right)\mathit{\boldsymbol{E}} \times \left( {\mathit{\boldsymbol{C}} + } \right.} \right.\\ \;\;\;\left. {\left. {{D_2}{K_2}\left( t \right)} \right)} \right] = 0,\\ \mathit{\boldsymbol{E'}}{P_1}\left( T \right)\mathit{\boldsymbol{E}} = {\mathit{\boldsymbol{H}}_1},\\ {K_1}\left( t \right) = - {\left( {{\mathit{\boldsymbol{R}}_{11}} + {{\mathit{\boldsymbol{D'}}}_1}\mathit{\boldsymbol{E'}}{P_1}\left( t \right)\mathit{\boldsymbol{E}}{\mathit{\boldsymbol{D}}_1}} \right)^{ - 1}} \times \\ \;\;\;\;\left[ {{{\mathit{\boldsymbol{B'}}}_1}{P_1}\left( t \right)\mathit{\boldsymbol{E}} + {{\mathit{\boldsymbol{D'}}}_1}\mathit{\boldsymbol{E'}}{P_1}\left( t \right)\mathit{\boldsymbol{E}}\left( {\mathit{\boldsymbol{C}} + {\mathit{\boldsymbol{D}}_2}{K_2}\left( t \right)} \right)} \right],\\ {\mathit{\boldsymbol{R}}_{11}} + {{\mathit{\boldsymbol{D'}}}_1}\mathit{\boldsymbol{E'}}{P_1}\left( t \right)\mathit{\boldsymbol{E}}{\mathit{\boldsymbol{D}}_1} > 0. \end{array} \right. $

(11)

$ \left\{ \begin{array}{l} \mathit{\boldsymbol{E'}}{{\dot P}_2}\left( t \right)\mathit{\boldsymbol{E}} = \mathit{\boldsymbol{E'}}{P_2}\left( t \right)\left( {\mathit{\boldsymbol{A}} + {\mathit{\boldsymbol{B}}_1}{K_1}\left( t \right)} \right) + \left( {\mathit{\boldsymbol{A}} + } \right.\\ \;\;\;\;{\left. {{\mathit{\boldsymbol{B}}_1}{K_1}\left( t \right)} \right)^\prime }{P_2}\left( t \right)\mathit{\boldsymbol{E}} + {\mathit{\boldsymbol{Q}}_2} + {{K'}_1}\left( t \right){\mathit{\boldsymbol{R}}_{21}}{K_1}\left( t \right) + \\ \;\;\;\;{\left( {\mathit{\boldsymbol{C}} + {\mathit{\boldsymbol{D}}_1}{K_1}\left( t \right)} \right)^\prime }\mathit{\boldsymbol{E'}}{P_2}\left( t \right)\mathit{\boldsymbol{E}}\left( {\mathit{\boldsymbol{C}} + {\mathit{\boldsymbol{D}}_1}{K_1}\left( t \right)} \right) + \\ \;\;\;\;{{K'}_2}\left( t \right)\left[ {{{\mathit{\boldsymbol{B'}}}_2}{P_2}\left( t \right)\mathit{\boldsymbol{E}} + {{\mathit{\boldsymbol{D'}}}_2}\mathit{\boldsymbol{E'}}{P_2}\left( t \right)\mathit{\boldsymbol{E}} \times \left( {\mathit{\boldsymbol{C}} + } \right.} \right.\\ \;\;\;\;\left. {\left. {{D_1}{K_1}\left( t \right)} \right)} \right] = 0,\\ \mathit{\boldsymbol{E'}}{P_2}\left( T \right)\mathit{\boldsymbol{E}} = {\mathit{\boldsymbol{H}}_2},\\ {K_2}\left( t \right) = - {\left( {{\mathit{\boldsymbol{R}}_{22}} + {{\mathit{\boldsymbol{D'}}}_2}\mathit{\boldsymbol{E'}}{P_2}\left( t \right)\mathit{\boldsymbol{E}}{\mathit{\boldsymbol{D}}_2}} \right)^{ - 1}} \times \\ \left[ {{{\mathit{\boldsymbol{B'}}}_2}{P_2}\left( t \right)\mathit{\boldsymbol{E}} + {{\mathit{\boldsymbol{D'}}}_2}\mathit{\boldsymbol{E'}}{P_2}\left( t \right)\mathit{\boldsymbol{E}}\left( {\mathit{\boldsymbol{C}} + {\mathit{\boldsymbol{D}}_1}{K_1}\left( t \right)} \right)} \right],\\ {\mathit{\boldsymbol{R}}_{22}} + {{\mathit{\boldsymbol{D'}}}_2}\mathit{\boldsymbol{E'}}{P_2}\left( t \right)\mathit{\boldsymbol{E}}{\mathit{\boldsymbol{D}}_2} > 0. \end{array} \right. $

(12)

假设系统(3)是均方容许的, 则

1) 有限时间随机奇异系统非零和博弈问题存在均衡解(u₁^*(t), u₂^*(t))，

$ u_1^ * \left( t \right) = {K_1}\left( t \right)x\left( t \right),u_2^ * \left( t \right) = {K_2}\left( t \right)x\left( t \right). $

(13)

2) 相应的最优值为

$ {J_\tau }\left( {u_1^ * ,u_2^ * ;{x_0},0} \right) = \frac{1}{2}{{x'}_0}\mathit{\boldsymbol{E'}}{P_\tau }\left( 0 \right)\mathit{\boldsymbol{E}}{x_0},\left( {\tau = 1,2} \right). $

(14)

证明  考虑最优控制问题(15)~(16)在K_i(t)= K_i^*(t)处取得最小值

$ \left\{ \begin{array}{l} \begin{array}{*{20}{c}} {\mathit{\boldsymbol{E}}{\rm{d}}x\left( t \right) = \left( {\mathit{\boldsymbol{A}} + {\mathit{\boldsymbol{B}}_j}K_j^ * \left( t \right) + {\mathit{\boldsymbol{B}}_i}{K_i}\left( t \right)} \right)x\left( t \right){\rm{d}}t + }\\ {\left( {\mathit{\boldsymbol{C}} + {\mathit{\boldsymbol{D}}_j}K_j^ * \left( t \right) + {\mathit{\boldsymbol{D}}_i}{K_i}\left( t \right)} \right)x\left( t \right){\rm{d}}w\left( t \right),} \end{array}\\ x\left( 0 \right) = {x_0},i,j = 1,2,i \ne j. \end{array} \right. $

(15)

$ \begin{array}{l} {J_i}\left( {u_i,u_j^ * ;{x_0},0} \right) = \varepsilon \left\{ {\frac{1}{2}\int_0^T {\left[ {x'\left( t \right)\left( {{Q_i} + } \right.} \right.} } \right.\\ \left. {\left. {K{{_j^ * }^\prime }\left( t \right){\mathit{\boldsymbol{R}}_{ij}}K_j^ * \left( t \right) + {{K'}_i}\left( t \right){\mathit{\boldsymbol{R}}_{ii}}{K_i}\left( t \right)} \right) \times x\left( t \right)} \right]{\rm{d}}t + \\ \left. {\frac{1}{2}x'\left( T \right){\mathit{\boldsymbol{H}}_i}x\left( T \right)} \right\}. \end{array} $

(16)

注意到最优控制问题(15)~(16)与引理1中的最优控制问题(6)~(7)相对应，笔者把引理1应用到最优控制问题(15)~(16)中，有如下的对应关系：

$\begin{array}{l} \mathit{\boldsymbol{A + }}{\mathit{\boldsymbol{B}}_\mathit{j}}\mathit{\boldsymbol{K}}_j^*\left( t \right) \Rightarrow \mathit{\boldsymbol{A, }}{\mathit{\boldsymbol{B}}_\mathit{i}} \Rightarrow \mathit{\boldsymbol{B}}, \mathit{\boldsymbol{C + }}{\mathit{\boldsymbol{D}}_\mathit{j}}\mathit{\boldsymbol{K}}_j^*\left( t \right) \Rightarrow \mathit{\boldsymbol{C}}\\ {\mathit{\boldsymbol{D}}_\mathit{i}} \Rightarrow \mathit{\boldsymbol{D, }}{\mathit{\boldsymbol{Q}}_\mathit{i}} + \mathit{\boldsymbol{K}}_j^*{'}\left( t \right){\mathit{\boldsymbol{R}}_{\mathit{ij}}}K_j^*\left( t \right) \Rightarrow \mathit{\boldsymbol{Q}}\\ {\mathit{\boldsymbol{R}}_{\mathit{ii}}} \Rightarrow \mathit{\boldsymbol{R, }}{\mathit{\boldsymbol{H}}_\mathit{i}} \Rightarrow \mathit{\boldsymbol{H}}\mathit{\boldsymbol{.}} \end{array} $

所以，最优控制策略存在

$ \begin{array}{l} {u^ * }\left( t \right) = - {\left( {\mathit{\boldsymbol{R}} + \mathit{\boldsymbol{D'E'}}P\left( t \right)\mathit{\boldsymbol{ED}}} \right)^{ - 1}} \times \left( {\mathit{\boldsymbol{B'}}P\left( t \right)\mathit{\boldsymbol{E}} + } \right.\\ \left. {\mathit{\boldsymbol{D'E'}}P\left( t \right)\mathit{\boldsymbol{EC}}} \right)x\left( t \right) \Rightarrow u_i^ * \left( t \right) = - {\left( {{\mathit{\boldsymbol{R}}_{ii}} + {{\mathit{\boldsymbol{D'}}}_i}\mathit{\boldsymbol{E'}}{P_i}\left( t \right)\mathit{\boldsymbol{E}}{\mathit{\boldsymbol{D}}_i}} \right)^{ - 1}} \times \\ \left[ {{{\mathit{\boldsymbol{B'}}}_i}{P_i}\left( t \right)\mathit{\boldsymbol{E}} + {{\mathit{\boldsymbol{D'}}}_i}\mathit{\boldsymbol{E'}}{P_i}\left( t \right)E\left( {\mathit{\boldsymbol{C}} + {\mathit{\boldsymbol{D}}_j}K_j^ * \left( t \right)} \right)} \right]x\left( t \right), \end{array} $

(17)

且相应的最优值为$\frac{1}{2}\mathit{x}{\mathit{'}_\mathit{0}}\mathit{\boldsymbol{E}}\mathit{'}{\mathit{P}_\mathit{i}}\left( 0 \right)\mathit{\boldsymbol{E}}{\mathit{x}_{\rm{0}}} $.

定理1 证毕.

5 结论

本文讨论了噪声依赖于状态与控制的随机奇异系统的非零和博弈问题，把随机奇异系统的线性二次最优控制的结果推广到两人非零和博弈问题，得到了有限时间随机奇异系统非零和博弈问题均衡解存在的充分条件等价于其相应耦合Riccati微分方程存在解，是对现有非零和博弈理论研究的丰富.然而，文章只是研究了噪声依赖于状态与控制的随机奇异系统的非零和博弈问题，而对应系统的零和博弈及主从博弈在现实决策中都有着重要的应用，因而相应的零和博弈、主从博弈等将有待进一步研究.